Difference between revisions of "2010 AMC 10B Problems/Problem 6"
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<math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65</math> | <math>\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65</math> | ||
− | ==Solution== | + | ==Solution 1== |
Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since <math>O</math> is the center, <math>OC</math> and <math>OA</math> are radii and they are congruent. Thus, <math>\triangle COA</math> is an isosceles triangle. Also, note that <math>\angle COB</math> and <math>\angle COA</math> are supplementary, then <math>\angle COA = 180 - 50 = 130^{\circ}</math>. Since <math>\triangle COA</math> is isosceles, then <math>\angle OCA \cong \angle OAC</math>. They also sum to <math>50^{\circ}</math>, so each angle is <math>\boxed{\textbf{(B)}\ 25}</math>. | Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since <math>O</math> is the center, <math>OC</math> and <math>OA</math> are radii and they are congruent. Thus, <math>\triangle COA</math> is an isosceles triangle. Also, note that <math>\angle COB</math> and <math>\angle COA</math> are supplementary, then <math>\angle COA = 180 - 50 = 130^{\circ}</math>. Since <math>\triangle COA</math> is isosceles, then <math>\angle OCA \cong \angle OAC</math>. They also sum to <math>50^{\circ}</math>, so each angle is <math>\boxed{\textbf{(B)}\ 25}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | An inscribed angle is always half its central angle, so therefore, half of 50 is 25, or B. | ||
+ | |||
+ | (Solution by Flamedragon) | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=5|num-a=7}} | {{AMC10 box|year=2010|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:10, 9 February 2014
Contents
Problem
A circle is centered at , $\overbar{AB}$ (Error compiling LaTeX. ! Undefined control sequence.) is a diameter and is a point on the circle with . What is the degree measure of ?
Solution 1
Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since is the center, and are radii and they are congruent. Thus, is an isosceles triangle. Also, note that and are supplementary, then . Since is isosceles, then . They also sum to , so each angle is .
Solution 2
An inscribed angle is always half its central angle, so therefore, half of 50 is 25, or B.
(Solution by Flamedragon)
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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